Yes and that follows by writing out another factorization for the number that divides $b = p_1 \cdots p_s$ and using unique factorization to dedude that the factors must be some of the primes in the factorization of $b$.
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Adrián BarqueroApr 7 '12 at 15:34

More explicitly, if $b = p_{1}^{e_1} \cdots p_{s}^{e_s}$ is the prime factorization of $b$ where the $p_i$'s are different primes, then you can prove that any divisor of $b$ is a product of the form $p_{1}^{a_1} \cdots p_{s}^{a_s}$ where the exponents $a_i$ satisfy that $0 \leq a_i \leq e_i$ for $i = 1, \dots , s$.
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Adrián BarqueroApr 7 '12 at 15:38

3 Answers
3

Essentially yes: If $p_1,\ldots,p_s$ are primes (possibly repeated), then every divisor of
$$b = p_1p_2\cdots p_s$$
must be a (possibly empty) product of some (possibly all) of the $p_i$, times $1$ or $-1$.

A divisor of $b$ has to be a product of some of the prime factors of $b$, but "some" may mean $0$ of them (if the divisor is $1$). The divisor can have no prime factors that are not prime factors if $b$.

If the divisor in question is $c$, then for some number $d$ we have $b=cd$. If a prime number $p$ divides $b$, then $p$ must divide either $c$ or $d$ (or both). The foregoing sentence is Euclid's lemma.

Hint $\rm\ p\ |\ d\ |\ p_1\cdots p_n\ \Rightarrow\ p\ |\ p_j\:$ for some $\rm\:j\:$ by the Prime Divisor Property. Cancelling $\rm\:p\:$ from $\rm\:d\:$ and $\rm\:p_1\cdots p_n\:$ and inducting yields that the prime factors of $\rm\:d\:$ are a sub-multiset of $\rm\{p_1,\ldots,p_n\}.$

Generally $\rm\:d\ |\ b_1\cdots b_n\ \Rightarrow\ d = d_1\cdots d_n,\ \ d_j\ |\ b_j,\:$ i.e. a divisor of a product is a product of divisors. This is a useful generalization of the Prime Divisor Property from atoms to composite numbers. Yours is the special case when all $\rm\:b_j = p_j\:$ prime, so $\rm\:d_j = 1\:$ or $\rm\:p_j\:$ (modulo a unit factor $\pm1$).